`n_(N_2)=\frac{44,8}{22,4}=2(mol)`
`n_(H_2)=\frac{18}{2}=9(mol)`
`n_(NH_3)=\frac{8,5}{17}=0,5(mol)`
$N_2+3H_2\xrightarrow{t^o}2NH_3$
`0,25` `0,75` `0,5`
`a,`
`V_(N_2)=22,4(2-0,25)=39,2(l)`
`V_(H_2)=22,4.(9-0,75)=184,8(g)`
`V_(NH_3)=0,5.22,4=11,2(l)`
`%V_(NH_3)=\frac{11,2}{39,2+184,8+11,2}.100=4,76%`
`%V_(N_2)=\frac{39,2}{235,2}.100=16,67%`
`%V_(H_2)=100-16,67-4,76=78,57%`
`b,`
`2NH_3+H_2SO_4->(NH_4)_2SO_4`
`0,5` `0,25`
`V_(H_2SO_4)=\frac{0,25}{2}=0,125(l)`